Series: convergent or divergent: E,n=1,10^n/(-9)^n-1

[icekitsune]

Hello,

I have two problems that i'm not too sure how to solve. I have to see if the series is convergent or divergent and if it is convergent, i have to find the sum.
1)

E 10^n/ (-9)^n-1
n=1

2)

E n+1 / 2n-3
n=1

i'm not sure how to start..thank you for the help!

 

[Deleted member 4993]

Re: Series: convergent or divergent

Hello,

I have two problems that i'm not too sure how to solve. I have to see if the series is convergent or divergent and if it is convergent, i have to find the sum.
1)

E 10^n/ (-9)^n-1
n=1

\(\displaystyle \sum_{n=1}^{\infty}\frac{10^n}{(-9)^{n-1}} \, = \, 10 \, + \, (\frac{10}{9})^2 + (\frac{10}{9})^4 ....\)


2)

E n+1 / 2n-3
n=1

(n+1)/(2n-3) = 1/2 * [1 + 5/(2n-3)] ............. Look at this series
i'm not sure how to start..thank you for the help!

Better yet check limit of the term as n ? infinity.

 

[BigGlenntheHeavy]

Both series diverge, the first one by r= -10/9, |r|>1, hence the series diverges.

For two, lim [a(sub)n] = 1/2, hence by nth-term test for divergence, it diverges.